Question

I was going through this to study copy constructor again.

CODE:

#include <iostream>

using namespace std;

class Line
{
   public:
      int getLength( void );
      Line( int len );             // simple constructor
      Line( const Line &obj);  // copy constructor
      ~Line();                     // destructor

   private:
      int *ptr;
};

// Member functions definitions including constructor
Line::Line(int len)
{
    cout << "Normal constructor allocating ptr" << endl;
    // allocate memory for the pointer;
    ptr = new int;
    *ptr = len;
}

Line::Line(const Line &obj)
{
    cout << "Copy constructor allocating ptr." << endl;
    ptr = new int;
   *ptr = *obj.ptr; // copy the value
}

Line::~Line(void)
{
    cout << "Freeing memory!" << endl;
    delete ptr;
}
int Line::getLength( void )
{
    return *ptr;
}

void display(Line obj)
{
   cout << "Length of line : " << obj.getLength() <<endl;
}

// Main function for the program
int main( )
{
   Line line(10);

   display(line);

   return 0;
}

OUTPUT:

Normal constructor allocating ptr
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!

I was wondering why it is calling Copy constructor in Line line(10). I think it should call the normal constructor only. I’m not cloning any object here. Please someone explain me this.

Answer 1

The copy constructor is called because the function parameter is passed by value. Here

void display(Line obj)

the function is defined as taking the Line parameter by value. This results in a copy of the argument being made. Had you passed by reference, the copy constructor would not be called